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A particle is moving in x-y plane. Its initial velocity and acceleration are $u = (4 \hat{i} + 8 \hat{j}) m / s$ and $a = (2 \hat{i} - 4 \hat{j}) m / s^{2}$ . Find the velocity of the particle t this instant.
Initial coordinates of particle are $(4 m, 10 m)$ .

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$(14 \hat{i} - 12 \hat{j}) m / s$

$(16 \hat{i} - 19 \hat{j}) m / s$

$(14 \hat{i} - 15 \hat{j}) m / s$

$(13 \hat{i} - 12 \hat{j}) m / s$

answer is A.

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Detailed Solution

A particle is moving in x-y plane. Its initial velocity and acceleration are u=(4 hati+8 hatj) m//s and a=(2 hati-4 hatj) m//s^2. Find (a) the time when the particle will cross the

y = 10 + 8t - $\frac{1}{2} 4 t^{2}$ = 0

Solving , t = 5 s

For constant acceleration , we can write

$v = u + a t v = (4 \hat{i} + 8 \hat{j}) + (2 \hat{i} - 4 \hat{j}) (5) = (14 \hat{i} - 12 \hat{j}) m / s$

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